Should the unbiased estimator of the variance of the sample proportion have (n-1) in the denominator?

Question

I know that the variance of the sample mean is:

$$\frac{\sigma^2}{n}$$

And that the unbiased estimator for that expression is:

$$\frac{\sigma^2}{n-1}$$

The variance of the sample proportion is:

$$\frac{p(1-p)}{n}$$

Does an unbiased estimator of variance of the sample proportion also need to have (n-1) in the denominator, like the unbiased estimator of the variance of the sample mean?

I feel the resoning for the denominator should be applicable for both situations, but I haven't seen it mentioned anywhere...

Answer 1

If $\hat{p}$ is the sample proportion, then $n\hat{p} \sim \text{Binomial}(n, p)$ so

$$E[\hat{p}(1-\hat{p})] = E[\hat{p}] - E[\hat{p}^2] = E[\hat{p}] - \text{Var}(\hat{p}) - E[\hat{p}]^2 = p(1-p) (1 - \frac{1}{n})$$ so it seems that the unbiased estimator of $\frac{p(1-p)}{n}$ would be $$\frac{\hat{p}(1-\hat{p})}{n-1}.$$

I might have made a miscalculation somewhere though.

Answer 2

If you have a random variable $X$ with a binomial distribution with parameters $n$ and $p$, then $\mathbb E\left[\frac{X}{n} \right]=p$, so the sample proportion $Y=\frac{X}{n}$ is an unbiased estimator of $p$.

Meanwhile $\mathrm{Var}\left(Y \right) = \mathrm{Var}\left(\frac{X}{n} \right) = \frac{p(1-p)}{n}$ is the variance of the sample proportion,

implying $\mathbb E[Y^2] = \frac{p(1-p)}{n}+\frac{p^2}{n^2}$,

so $\mathbb E\left[Y(1-Y) \right] = p - \frac{p(1-p)}{n}-\frac{p^2}{n^2} = p(1-p)\frac{n-1}{n} $ and thus $\mathbb E\left[\frac{Y(1-Y)}{n-1} \right] = \frac{p(1-p)}{n}$,

showing $\frac{Y(1-Y)}{n-1} $ is an unbiased estimator of the variance of the sample proportion $\frac{p(1-p)}{n}$